Question

HELP ASAP: Determine the enthalpy change of the following reaction: CH4 + 4Cl2 -> CCl4+ 4HCl 

Given enthalpies: 
CH4: -17.9 kJ/mol 
Cl2: 0 kJ/mol 
CCl4: -95.98 kJ/mol 
HCl: -92.3 kJ/mol

A. -447.28 kJ/mol

B. -206.18

C. -170.38

D. -393.58​

Answer 1

Answer: A. -447.28

Step-by-step explanation:

AI-generated answerTo determine the enthalpy change of a reaction, we need to calculate the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants. The given reaction is: CH4 + 4Cl2 -> CCl4 + 4HCl Given enthalpies: CH4: -17.9 kJ/mol Cl2: 0 kJ/mol CCl4: -95.98 kJ/mol HCl: -92.3 kJ/mol We can calculate the enthalpy change as follows: Enthalpy change = (Sum of enthalpies of products) - (Sum of enthalpies of reactants) Enthalpy change = [(-95.98 kJ/mol) + (4 * -92.3 kJ/mol)] - [(-17.9 kJ/mol) + (4 * 0 kJ/mol)] Enthalpy change = [-95.98 kJ/mol - 369.2 kJ/mol] - [-17.9 kJ/mol] Enthalpy change = -465.18 kJ/mol + 17.9 kJ/mol Enthalpy change ≈ -447.28 kJ/mol Therefore, the correct answer is C. -447.28 kJ/mol.

Answer 2

A

Step-by-step explanation:

In this question, we are to calculate the enthalpy of change of the reaction. ΔH

To be able to do that, we use the Hess’ law and it involves the subtraction of the summed heat reaction of the reactants from that of the product.

Thus, mathematically, the enthalpy of change of the reaction would be;

[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4) + 4 ΔH(Cl2)]

We can see that we multiplied some heat change by some numbers. This is corresponding to the number of moles of that compound in question in the reaction.

Also, for diatomic gases such as chlorine in the reaction above, the heat of reaction is zero.

Thus, we can have the modified equation as follows;

[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4)]

Substituting the values we have according to the question, we have;

-95.98 + 4(-92.3) -(-17.9)

= -95.98 - 369.2 + 17.9

= -447.28 KJ/mol